Calculating a limit of a sequence of integrals

Question

I have the following homework problem:

Let $\mathcal{B}(\mathbb{R})$ be the Borel $\sigma$-algebra on $\mathbb{R}$ and $\lambda$ the Lebesgue-measure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$. Define for every $n\geq 1$ the following function $f_{n}: \mathbb{R} \to \mathbb{R}$ with $f_{n}(x)= \frac{\sin(n^{2}x^{2}}{n^{2}|x|^{3/2}} 1_{(-n,n)\setminus \{0\} }(x)$. Determine $$ \lim\limits_{n\to\infty} \int_{\mathbb{R}} f_{n} d\lambda $$

I had attempted the following:

First I wrote out that the function $f_{n}(x)$ is continous and therefore measurable. Then I tried to estimate $|f_{n}(x)|$ so that I can apply the Dominated convergence theorem. I initially tried the function $w(x)=\frac{1}{|x|^{3/2}}$ which diverges.

The other function that I tried is $w(x)=\sqrt{|x|}$ which does bound the entire function but I can't calculate the riemann integral of this function.

Since this didn't work out I tried to take the $\lim\limits_{n\to\infty} f_{n}(x)=0$. But then I'm unsure which w(x) to take as the constant function $w(x)=0$ doesn't work because $|f_{n}(x)|\geq 0$

So I'm not entirely sure which $w(x)$ to take to apply the Dominated Convergence theorem from which I can then determine the integral.

Answer 1

Take $w(x)=1_{(-1,1)}+\frac 1 {|x|^{3/2}}1_{|x|>1}$. [For $|x|<1$ use the fact that $\sin t \leq t$ for $t >0$].

Answer 2

Notice that

$$I_n=\int_{[-n,n]} f_n=\int_{\{|x|\leq 1\}}f_n\,dx+ \int_{\{|x|>1\}}f_n\,dx$$

On $\{|x|\leq1\}$, $$|f_n(x)|=\Big|\frac{\sin (nx)^2}{(nx)^2}\Big||x|^{1/2}\leq|x|^{1/2}$$

On $\{|x|>1\}$, $$|f_n(x)|=\Big|\frac{\sin (nx)^2}{n^2 |x|^{3/2}}\Big|\leq\frac{1}{|x|^{3/2}}$$

That is $$|f_n(x)|\leq |x|^{1/2}\mathbb{1}_{\{|x|\leq1\}}+\frac{1}{|x|^{3/2}}\mathbb{1}_{\{|x|>1\}}\in L_1(\mathbb{R})$$

One can then apply dominated convergence. Observe that $\lim_nf_n(x)=0$ poitwise.